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yours question is not correct

no it is a correct question

we have to prose ure answer

*yours

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sec x dx = sec x sec x - tan x sec x - tan x dxset

u = sec x - tan x

then we find

du = (sec x tan x - sec2 x) dxsubstitute du = (sec x tan x - sec2 x) dx, u = sec x - tan x

sec x sec x - tan x sec x - tan xdx = (sec2 x - sec x tan x) dx sec x - tan x = du usolve integral= ln |u| - Csubstitute back u=sec x - tan x= ln |sec x - tan x| + C